Is the function given below continuous/differentiable at $x=0$ ? $h(x)=\begin{cases} 2x^2- x&,&x<0 \\\\ -x&,&x\geq0 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Explanation: Checking for continuity at $x=0$ For the function to be continuous at $x=0$, we need the two-sided limit $\lim_{x\to 0}h(x)$ to exist and be equal to $h(0)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 0^-}h(x)$ and $\lim_{x\to 0^+}h(x)$ exist and are equal to $h(0)$. According to $h$ 's definition, $h(0)=-(0)=0$. $\lim_{x\to0^-}h(x)$ $2x^2-x$ evaluated at $x=0$ is equal to $0$. Since $2x^2-x$ is continuous, we can be certain that $\lim_{x\to 0^-}h(x)=0$. $\lim_{x\to 0^+}h(x)$ $-x$ evaluated at $x=0$ is equal to $0$. Since $-x$ is continuous, we can be certain that $\lim_{x\to 0^+}h(x)=0$. We saw that the two one-sided limits exist and are equal to $h(0)$, so the function is continuous at $x=0$. Checking for differentiability at $x=0$ For the function to be differentiable at $x=0$, we need the two-sided limit $\lim_{x\to 0}\dfrac{h(x)-h(0)}{x-0}=\,\lim_{x\to 0}\dfrac{h(x)-0}{x-0}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 0^-}\dfrac{h(x)-0}{x-0}$ and $\lim_{x\to 0^+}\dfrac{h(x)-0}{x-0}$ exist and have the same value. $\lim_{x\to 0^-}\dfrac{h(x)-0}{x-0}=-1$ $\lim_{x\to 0^+}\dfrac{h(x)-0}{x-0}=-1$ We saw that the two one-sided limits exist and are equal, so the function is differentiable at $x=0$. In conclusion, the function is both continuous and differentiable at $x=0$.